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Chris gave Brian a four-digit code to disarm Chris’s security system.

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Intern
Joined: 07 Aug 2016
Posts: 21
Location: India
WE: Consulting (Energy and Utilities)
Chris gave Brian a four-digit code to disarm Chris’s security system.  [#permalink]

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Updated on: 24 Apr 2017, 00:39
10
00:00

Difficulty:

55% (hard)

Question Stats:

55% (01:34) correct 45% (01:36) wrong based on 108 sessions

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Chris gave Brian a four-digit code to disarm Chris’s security system. However, when Brian typed in the code, it didn’t work. If Brian realizes that Chris must have written exactly one digit incorrectly, but Brian does not know which digit is incorrect or what its correct value is, how many more attempts will Brian need to guarantee that he identifies the correct four-digit code?

A. 36
B. 39
C. 3024
D. 6561
E. 9999

My analysis was -

"Since, does not know which digit is incorrect or what its correct value is" there are 10 possibilities for each digit of the code (0, 1, 2, 3, 4, 5, 6, 7, 8, 9). However, since it is 4-digit code, 0 cant be a possibility for the 1 digit of the code. so the number of attempt would be --
4*10 -1 = 39.

Can someone please explain why I am wrong !!

Originally posted by RudraM on 24 Apr 2017, 00:12.
Last edited by Bunuel on 24 Apr 2017, 00:39, edited 1 time in total.
Renamed the topic.
Senior Manager
Joined: 24 Apr 2016
Posts: 327
Re: combination problem !!  [#permalink]

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24 Apr 2017, 00:46
Let us say Chris gave Brian the Code: 1234

Now it says only one of the digits may be wrong. This means either 1,2,3,or 4 is incorrect.

Starting with the 1st digit '1'. If the first digit is not 1, then it can take 9 other values (0,2,3,4,5,6,7,8,9).
Note: The question does not state that 0 cannot be the first number of the code. If the question was asking about a 4 digit number, then yes, 0 cannot be the first number.But a code can definitely take 0 as the first digit and moreover the question does not talk about any such limitation.

Similarly the other three digits can also take 9 other possible numbers. Hence the total number of try outs Brian has to do is 9*4 = 36.

Let me know if this makes sense?
Senior Manager
Joined: 13 Oct 2016
Posts: 363
GPA: 3.98
Re: combination problem !!  [#permalink]

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24 Apr 2017, 00:48
1
3
RudraM wrote:
Chris gave Brian a four-digit code to disarm Chris’s security system. However, when Brian typed in the code, it didn’t work. If Brian realizes that Chris must have written exactly one digit incorrectly, but Brian does not know which digit is incorrect or what its correct value is, how many more attempts will Brian need to guarantee that he identifies the correct four-digit code?

A. 36
B. 39
C. 3024
D. 6561
E. 9999

My analysis was -

"Since, does not know which digit is incorrect or what its correct value is" there are 10 possibilities for each digit of the code (0, 1, 2, 3, 4, 5, 6, 7, 8, 9). However, since it is 4-digit code, 0 cant be a possibility for the 1 digit of the code. so the number of attempt would be --
4*10 -1 = 39.

Can someone please explain why I am wrong !!

Hi

My thoughts.

First of all, we are talking about 4 digit code not about four digit number, that means first digit can be zero. (0459 is also four digit code).

Now let's take our code:

A B C D

We know for sure that only one digit is incorrect, others are correct.

Let's assume that A is incorrect. We already typed 1 digit and it does not work, hence we are left with 9 digits to try. Total:

9*1*1*1 = 9 choices for incorrect A. If it still does not work, that means A is correct and we are moving to B.

Now we are assuming that B is not correct. Same logic:

1*9*1*1 = 9 choices.

Worst scenario will be when the correct digit appears to be the last we try on D. Hence we need to try total:

9*4 = 36 times.

Answer A.

Hope this helps.

Cheers.
Manager
Joined: 24 Sep 2018
Posts: 140
Chris gave Brian a four-digit code to disarm Chris’s security system.  [#permalink]

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24 Sep 2018, 23:15
1
Quote:
Chris gave Brian a four-digit code to disarm Chris’s security system. However, when Brian typed in the code, it didn’t work. If Brian realizes that Chris must have written exactly one digit incorrectly, but Brian does not know which digit is incorrect or what its correct value is, how many more attempts will Brian need to guarantee that he identifies the correct four-digit code?

A. 36
B. 39
C. 3024
D. 6561
E. 9999
The trick to this combinatorics problem is realizing that you want "or" (add) rather than "and" (multiply).

Brian needs to fix the first digit (9 possibilities if the given first digit was wrong) or the second digit (9 more possibilities) or the third digit (9 more) or the fourth digit (9 more).

The worst case scenario is that the very last digit is the one that needed correction.

Therefore the answer is 9 + 9 + 9 + 9 = 36, and A is correct.
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Chris gave Brian a four-digit code to disarm Chris’s security system.   [#permalink] 24 Sep 2018, 23:15
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